$f(x)=\sum_{n=0}^{\infty }\dfrac{n+1}{4^{n+1}}x^n$ $\int\limits_{0}^{1}{f(x)dx}=$
Solution: First, perform the requested integration. $\begin{aligned} &\phantom{=}\int\limits_{0}^{1}{f\left( x \right)dx} \\\\ &=\int_0^1\sum_{n=0}^{\infty }\dfrac{n+1}{4^{n+1}}x^ndx \\\\ &=\sum_{n=0}^{\infty }\int_0^1\dfrac{n+1}{4^{n+1}}x^ndx \\\\ &=\left. \sum_{n=0}^{\infty }\dfrac{n+1}{4^{n+1}}\dfrac{x^{n+1}}{n+1}\right|_{0}^{1} \\\\ &=\sum_{n=0}^{\infty }{\dfrac{1}{{{4}^{n+1}}}} \end{aligned}$ Now recognize that the result is a geometric series with first term $\dfrac{1}{4}$ and common ratio $\dfrac{1}{4}$. Hence, the series converges to $\dfrac{\dfrac{1}{4}}{1-\dfrac{1}{4}}=\dfrac{1}{4-1}=\dfrac{1}{3}$.